Electrostatics Note

1

Static Electric Fields Electrostatics is the study of the effects of electric charges at rest, and the static electric fields,

which are caused by stationary electric charges.

In the deductive approach, few fundamental relations for an idealized model are postulated as

axioms, from which particular laws and theorems can be derived. Then the validity of the

model and the axioms are verified by the experiments. The steps involved in building a theory

based on an idealized model are as follows:

1. Define some basic quantities. (E, q)

2. Specify the rules of operations. (Vector analysis)

3. Postulate some fundamental relations. (Divergence equation, Curl equation)

3-2 Electrostatics in Free Space

Here, electric field in free space is considered. Noted that the permittivity1 of the free space,

denoted by ε0, is equal to (1/36π)×10-9

= 8.854×10-12

(F/m).

First, electric field intensity is defined as the force per unit charge that a very small

stationary test charge experiences when it is placed in a region where an electric field exists.

That is,

qq

FE

0lim→

= (V/m) (3-1)

Thus, E is proportional to and in the direction of the force F. Notice that the unit

Newton/Coulomb = V/m. An inverse relation of (3-1) gives

EF q= (N) (3-2)

The two fundamental postulates of electrostatics in free space specify the divergence and the

curl of E. They are

0ερv=⋅∇ E (3-3)

and

0E =×∇ , (3-4)

where ρv denotes the volume charge density with the unit (C/m3). The definition of ρv is given

by

v

q

vv ∆

∆=

→∆ 0limρ (C/m

3)

(3-4) asserts that static electric fields are irrotational whereas (3-3) implies that a static

electric field is not solenoidal. These two equations are point relations or in differential forms.

Taking the volume integral of both sides of (3-3) over a volume V yields

00 εερ Q

dvdvV

v

V==⋅∇ ∫∫ E

where Q is the total charge contained in V. Applying the divergence theorem, one obtains

0εQ

dS

=⋅∫ sE (Gauss’s law) (3-6)

which is a form of Gauss’s law. Likewise, taking the surface integral of both sides of (3-4)

and applying Stokes’ theorem yields

0=⋅∫C dllllE , (3-7)

1

Permittivity is a physical quantity that describes how an electric field affects, and is affected by, a

dielectric medium, and is determined by the ability of a material to polarize in response to the field, and

thereby reduce the total electric field inside the material. Thus, permittivity relates to a material's ability

to transmit (or "permit") an electric field.

Electrostatics Note

2

which asserts that “the scalar line integral of the static electric field intensity around any

closed path vanishes”. (3-6), (3-7) are of integral forms. Since the scalar product E·dl

integrated over any path is the voltage along that path, i.e.,

∫ ⋅=C

dV lE (V),

thus (3-7) is equivalent to Kirchhoff’s voltage law, i.e., the algebraic sum of voltage drops

around any closed circuit is zero.

3-3 Coulomb’s Law Consider a single point charge q at rest in boundless free space. In order to find the electric

field intensity due to q, a spherical surface of an arbitrary radius r centered at qa

hypothetical enclosed surface (a Gaussian surface) around the source is drawn, upon which

Gauss’s law is applied to determine the field. Since a point charge has no preferred directions,

its electric field must be everywhere radial and has the same intensity at all points on the

spherical surface. Applying (3-6) to Fig. 1 (a) yields

0

ˆ)ˆ(εq

dsEdS

rS

=⋅=⋅ ∫∫ rrsE or 0

24ε

πq

rEdsE rS

r ==∫ .

Therefore,

2

04ˆˆ

r

qEr πε

rrE == (V/m) (3-8)

Fig. 1

From (3-8), the electric field intensity of a point charge is in the outward radial direction and

has a magnitude proportional to the charge and inversely proportional to the square of the

distance from the charge. If the charge q is not located at the origin, referring to Fig. 1(b), one

obtains the electric field intensity at point P to be

2

0 |'|4ˆ

rrE

−=

πεq

aqPP .

But |'|

'ˆ

rr

rr

−−

=qPa

3

0 |'|4

)'(

rr

rrE

−−

=πεq

P (V/m) (3-11)

Example 3-1 Determine the electric field intensity at P(-0.2,0,-2.3) due to a point charge of

+5 (nC) at Q(0.2,0.1,-2.5) in air. All dimensions are in meters.

Electrostatics Note

3

When a point charge q2 is placed in the electric field of another point charge q1, a force F12 is

experienced by q2 due to E12 of q1 at q2, which is given by

2

120

211212212

4ˆ

R

qqaq

πε== EF (N) ; |||| 121212 rrR −==R (3-13)

(3-13) is a mathematical form of Coulomb’s law : the force between two point charges is

proportional to the product of the charges and inversely proportional to the square of the

distance of separation.

Example 3-2 The electrostatic deflection

system of a cathode-ray oscillograph is

depicted in the right figure. Electrons from

a heated cathode are given an initial

velocity 00ˆuzu = by a positively charged

anode. The electrons enter at z=0 into a

region of deflection plates where a uniform electric field dd EyE ˆ−= is maintained over a width w. Ignoring

gravitational effects, find the vertical deflection of the electrons on the fluorescent screen at

z=L.

3-3.1 Electric field due to a system of discrete charges Suppose an electrostatic field is created by a group of n discrete point charges, q1, q2, …, qn,

located at different positions, the principle of superposition can be applied to find the total

electric field due to this system of discrete charges, which is given by

3

0

2

3

20

22

3

10

11

||4

)(

||4

)(

||4

)(

n

nqqq

rr

rr

rr

rr

rr

rrE

′−

′−++

′−

′−+

′−

′−=

πεπεπεL or

∑= ′−

′−=

n

k k

kkq

13

0 ||

)(

4

1

rr

rrE

πε (V/m) (3-14)

3-3.2 Electric field due to a continuous distribution charges The electric field caused by a continuous distribution of charge as shown in the figure on the

right can be obtained by integrating the contribution of

an element of charge over the charge distribution. Let

ρv be the volume charge density (C/m3), then the

electric field intensity due to qdv’ at P is given by

2

04

'ˆ

R

dvad v

R περ

=E

Therefore,

∫∫∫ ===' 3

0' 2

0'

'4

1'ˆ

4

1

Vv

V

vR

Vdv

RR

dvad ρ

περ

πεR

EE (V/m) ; R

aR

R=ˆ (3-16)

Electrostatics Note

4

For the charge distributed on a surface with a surface charge density ρs (C/m2) (3-16)

becomes

∫=' 2

0

'ˆ

4

1

S

sR

R

dsa

ρπε

E (V/m) (3-17)

For a line charge with a line charge density ρl (C/m), (3-16) becomes

∫=' 2

0

'ˆ

4

1

LR

R

da

ll

ρπε

E (V/m) (3-18)

Example 3-3 Determine the electric field of an infinitely long,

straight, line charge of uniform density ρl (C/m) in air

3-4 Gauss’s Law and Applications Gauss’s law follows directly from (3-3) and is given by

0ε

Qd

S=⋅∫ sE (3-6)

Gauss’s law asserts that the total outward flux of the E-field over any closed surface in free

space is equal to the total charge enclosed in the surface divided by ε0. The surface S can be

hypothetical closed surface chosen for convenience, not necessarily be a physical surface.

Gauss’s law is useful in determining E when the normal component of the electric field

intensity is constant over an enclosed surface. The first step to apply Gauss’s law is to choose

such surface, referred to as a Gaussian surface, and then evaluate both sides of (3-6) in order

to determine E.

Example 3-4 Use Gauss’s law for Example 3-3

Electrostatics Note

5

Example 3-5 Determine the electric field

intensity due to an infinite planar charge

with a uniform surface charge density ρs.

Example 3-6 Determine the E field due to a spherical cloud

of electrons with a volume charge density ρ0 inside and 0

outside.

3-5 Electric Potential

Since ( ) 0≡∇×∇ V and 0E =×∇ in electrostatics, one can define a scalar electric potential

V from (3-4) such that

V−∇=E (3-26)

Electric potential is related to the work in carrying a charge from one point to another. Since

the electric field intensity is the force acting on a unit test charge, the work required to move a

unit charge from point P1 to P2 is given by

∫ ⋅−=2

1

P

Pd

q

WllllE (J/C or V) (3-27)

Since the static electric field is “conservative”, the line

integral on the right does not depend on the integration

path, for instance integrations along path 1 and path 2

give the same result.

Analogous to the concept of potential energy in mechanics, (3-27) represents the difference in

electric potential energy of a unit charge between point P2 and point P1. Let V denote the

electric potential energy per unit charge, the electric potential, then

∫ ⋅−=−2

112

P

PdVV llllE (V) (3-28)

since 12

2

1

2

1

2

1

ˆ VVdVdVdP

P

P

Pl

P

P−==⋅∇=⋅− ∫∫∫ laE llll . Thus, a potential difference (electrostatic

voltage) is equivalent to the electric potential energy per unit charge. Note that point P1 here

is the reference zero-potential point. In most cases, the reference point is taken at infinity; this

convention normally applies when the reference point is not specified explicitly.

Observations regarding electric potential

1. Because of the negative sign, the direction of E is opposite to the direction of increasing

V.

Electrostatics Note

6

2. The direction of V∇ is normal to

surfaces of constant V, thus E is

perpendicular to equipotential lines or

equipotential surfaces.

3-5.1 Electric Potential due to a charge distribution Let infinity be the reference point, then the electric potential of a point at a distance R from a

point charge q is given by

R

qdr

r

qV

R

0

2

0 4ˆ

4ˆ

πεπε=⋅−= ∫∞ rr (V) (3-29)

The potential difference between 2 points, P2 and P1, at distances R2 and R1, respectively, is

given by

−=−=

120

21

11

412 RR

qVVV PP πε

(3-30)

The electric potential due to a system of n discrete charges, q1, …, qn, is given by

∑= ′−

=n

k k

kqV

10 ||4

1

rrπε (V) (3-31)

For continuous charge distributions in confined regions, electric potentials are given by

∫='

0

'

4

1

V

v

R

dvV

ρπε

(V/m) (volume charge) (3-38)

∫='

0

'

4

1

S

s

R

dsV

ρπε

(V) (surface charge) (3-39)

∫='

0

'

4

1

L R

dV

ll

ρπε

(V) (line charge) (3-40)

Example 3-7 [Electric dipole moment] Electric potential due to an electric dipole consisting

of charges +q and –q with a small separation of d (assume R >> d)

Example 3-8 Obtain a formula for the electric field intensity on the axis of a circular disk of

radius b that carries a uniform surface charge density ρs.

Electrostatics Note

7

3-6 Material Media in Static Electric Field Consider energy band theory of solids based on solid state physics as shown in Fig. 2,

electrical materials can be classified into 3 types, namely, conductors, dielectrics2 (or

insulators), and semiconductors.

Figure 2: Energy band structure

3-6.1 Conductors in Static Electric Field

Assume that some electric charges are introduced in the interior of a good conductor. An

electric field will be set up and create a force that causes the movement of charges. This

movement will continue until all charges reach the conductor surface and redistribute in such

a way that both the charge and the field inside vanish. Hence,

0E == ;0vρ

When there are no free charges in the interior of a conductor ( 0=vρ ), E must be zero

according to Gauss’s law. Furthermore, under static conditions the E field on a conductor

surface is everywhere normal to the surface, otherwise there exists a tangential force that

moves the charges.

Consider the boundary conditions at the interface between a conductor and free space as

shown in Fig. 3.

Figure 3: A conductor-free space interface

Integrating E along the contour abcda and

taking the limit as 0→∆h yield

0

or0lim0

=

=∆=⋅∫→∆

t

tabcdah

E

wEdllllE

Which says that the tangential component of

the E field on a conductor surface is zero

under static conditions. In other words, the

surface of a conductor is an equipotential

surface.

Next, integrating E on the Gaussian surface in

the figure and taking the limit as 0→∆h :

00

orερ

ερ s

ns

nS

ESSEd =∆=∆=⋅∫ sE

2 A dielectric is a nonconducting substance, i.e. an insulator. The term was coined by William Whewell in

response to a request from Michael Faraday. Although "dielectric" and "insulator" are generally considered

synonymous, the term "dielectric" is more often used to describe materials where the dielectric polarization is

important, such as the insulating material between the metallic plates of a capacitor, while "insulator" is more often

used when the material is being used to prevent a current flow across it.

Electrostatics Note

8

Example 3-9 A positive point charge Q is at the center of a spherical conducting shell of an

inner radius Ri and an outer radius Ro. Determine E and V as functions of the radial distance r.

Figure 3: Example 3-9

3-6.2 Dielectrics in Static Electric Field

All material media are composed of atoms with a positively charged nucleus surrounded by

negatively charged electrons. In the absence of an external electric field, the molecules of

dielectrics are macroscopically neutral. The presence of an electric field causes a force on

each charged particle and results in small displacements of positive and negative charges in

opposite directions. These are bound charges. The displacements polarize a dielectric

material and create electric dipoles (i.e., polarization). The molecules of some dielectrics

possess permanent dipole moments, even in the absence of an external electric field. Such

molecules are called polar molecules, in contrast to nonpolar molecules. An example is the

water molecule H2O. Generally, dielectric materials consist of both polar and nonpolar

molecules (Fig. 5).

When there is no external field, dipoles in polar dielectrics are randomly oriented

(Fig. 6 (a)), producing no net dipole moment macroscopically. An applied electric field will

tend to align the dipoles with the field as shown in Fig. 6 (b). producing the nonzero net

dipole moment (Fig. 7).

Fig. 5: Molecules in dielectrics

Electrostatics Note

9

Figure 6: Polar molecule Figure 7: Interior of a dielectric medium

A polarization vector P is defined as

)(C/mlim 21

0 v

vN

k

k

v ∆=

∑∆

=

→∆

p

P

where N is the number of molecules per unit volume and the numerator represents the vectopr

sum of the induced dipole moments contained in a very small volume ∆v. The vector P is the

volume density of electric dipole moment. The dipole moment dp produces an electric

potential

'4

ˆ2

0

dvR

dVπε

rP ⋅=

Thus, the potential due to the polarized dielectric is given by

∫⋅

=' 2

0

'ˆ

4

1

Vdv

RV

rP

πε

Interpretation of the effects of the induced electric dipoles:

1. Equivalent polarization surface charge density

nps aP ˆ⋅=ρ (C/m2)

2. Equivalent polarization volume charge density Since

( ) ∫∫∫ =⋅∇−=⋅−=V

pvVS

n dvdvdsQ ρPaP ˆ ,

Thus, one can define the polarization volume charge density as

P⋅−∇=pvρ (C/m3)

It follows that

( ) 0ˆcharge total =⋅∇−⋅=+= ∫∫∫∫ VSn

Vpv

Sps dvdsdvds PaPρρ ,

i.e., the total “free” charge of the dielectric body after polarization must remain zero.

Example 3-10 The polarization vector in a dielectric sphere of radius R0 is 0ˆPxP = .

Determine

a) the equivalent polarization surface and volume charge densities and

b) the total equivalent charge on the surface and inside of the sphere

Electrostatics Note

10

3-7 Electric Flux Density and Dielectric Constant In dielectrics,

( )pvv ρρ

ε+=⋅∇

0

1E ,

but since P⋅−∇=pvρ , ( ) vρε =+⋅∇ PE0 .

Here, one can define a new fundamental field quantity, Electric Flux Density (or electric

displacement) D to be

PED += 0ε (C/m2)

It follows that

vρ=⋅∇ D

Applying the divergence theorem yields

QdvddvV

vV

==⋅=⋅∇ ∫∫∫ ρS

sDD .

Hence, Qd =⋅∫S sD (C)

In linear and isotropic media, P can be given in terms of E as

EP eχε 0=

where eχ is called electric susceptibility (dimensionless). Here, D can be rewritten as

( ) EEED εεεχε ==+= re 00 1 (C/m2)

where

0

1εε

χε =+= er

is called relative permittivity or dielectric constant (dimensionless).

In general, dielectric materials can be classified based on the property of dielectric constants

into

Linear : dielectric constant doesn’t change with applied electric field ↔ non-linear

Isotropic: dielectric constant doesn’t change with direction ↔ anisotropic

Homogeneous: dielectric constant doesn’t change from point to point ↔ inhomogeneous

3-7.1 Dielectric Strength

If the electric field is very strong, it will pull electrons completely out of molecules. The

electrons will accelerate under the influence of the electric field, collide violently with the

molecular structure and avalanche effect of ionization due to collisions may occur. The

material will become conducting and large currents may result; this phenomenon is called a

dielectric breakdown. The maximum electric field intensity that a dielectric material can

withstand without breakdown is the dielectric strength. For instance, the dielectric strength of

air at the atmospheric pressure is 3 (kV/mm).

Example 3-11 Consider two spherical conductors with radii b1 and b2 (b1 > b2) that are

connected by a conducting wire. The distance of separation between the conductors is

assumed to be very large in comparison to b2 so that the charges on the spherical conductors

may be considered as uniformly distributed.

a) the charges on the two spheres, and

b) the electric field intensities at the sphere surfaces.

Electrostatics Note

11

Example 3-9* A positive point charge Q is at the center of a spherical dielectric shell, with a

dielectric constant of εr and (inner, outer) radii, (Ri,Ro), respectively. Determine E, V, D, P as

functions of the radial distance r.

Figure 8 : Example 3-9*

3-8 Boundary Conditions for Electrostatic Fields

Consider the boundary conditions at the interface between two dielectric media as shown in

Fig. 9.

Integrating E along the contour abcda and taking the limit as 0→∆h yield

ttabcdah

EEEEd 21210

or 0)(lim ==∆−+∆=⋅∫→∆wwlE

which says that the tangential component of the E

Electrostatics Note

12

field is continuous across the interface. If ε1, ε2

denote the permittivities of media 1, 2, respectively,

then

2

2

1

1

εεtt DD

=

Next, integrating E on the Gaussian surface in the

figure and taking the limit as 0→∆h :

( )( ) SS

Sd

sn

nnS

∆=∆−⋅=

∆⋅+⋅=⋅∫ρ212

1221

ˆ

ˆˆ

DDa

aDaDsD

Figure 9 : An interface between two

media

Thus,

( ) snnsn DD ρρ =−=−⋅ 21212 orˆ DDa (C/m2)

where ρs denotes the surface charge density on the interface.

Example 3-13 A lucite sheet (εr=3.2) is introduced perpendicularly in a uniform electric field

0ˆEo xE = in free space. Determine Ei,Di,Pi inside the lucite.

Example 3-14 Two dielectric media with permittivities ε1 and ε2 are separated by a charge

free boundary. The electric field intensity in medium 1 at the point P1 has a magnitude E1 and

makes an angle α1 with the normal. Determine the magnitude and direction of E at point P2 in

medium 2.

Electrostatics Note

13

3-9 CAPACITANCES AND CAPACITORS

It is known from 3-6 that a conductor in a static electric field is an equipotential body

and that charges on a conductor will distribute themselves in such a way that the

electric field inside vanishes. Suppose the potential due to a charge Q is V, then

increasing the total charge by a factor k only increases the surface charge density

without changing the charge distribution. It is also noted that increasing Q also leads

to increasing E and thus V also increases. Reciprocally, increasing V by a factor of k

leads to increase in Q.

∫ ∫−=⋅−= dndV s

0ερ

llllE and ∫ ∫−=⋅−= dnk

dkkV s

0ερ

llllE ,

Thus, one can conclude that the Q/V ratio remains unchanged. This ratio is called the

capacitance of the isolated conducting body, which has the unit Farad (F), or C/V.

Using C, one can write

CVQ =

Of considerable importance in practice is the Capacitor (or Condenser) as shown in

Fig. 10.

Figure 10 : A two-conductor capacitor

Here, the capacitor consists of two

conductors separated by free space or a

dielectric medium. When a dc voltage

source is applied between conductors, a

charge transfer occurs, resulting in +Q on

one conductor and –Q on the other. Note

that the field lines are perpendicular to the

conductor surfaces. Let V12 be the

potential difference between two

conductors, then the capacitance C is

given by

12V

QC = (F)

The capacitance of a capacitor depends on

the geometry and the permittivity of the

medium.

Example 3-15 A parallel-plate capacitor consists of two parallel conducting plates of

areas S separated by a uniform distance d. The space between the plates is filled with

a dielectric of a constant permittivity ε. Determine the capacitance.

Example 3-16 A cylindrical capacitor

Electrostatics Note

14

Example 3-16* A spherical capacitor

3-10 ELECTROSTATIC ENERGY AND FORCES

Since electric potential at a point in an electric field is the work required to bring a

unit charge from infinity (the reference point) to that point, to bring a charge Q2 from

infinity against the field of a charge Q1 in free space to a distance R12 requires the

work of amount

11

120

21

120

12222

44VQ

R

QQ

R

QQVQW ====

πεπε,

which is path-independent. The work is stored in the assembly of two charges as

potential energy,

( )221122

1VQVQW +=

Here, if a charge Q3 is brought from infinity to a point that is R13 from Q1 and R23

from Q2, then an additional amount of work is required that equals

+==∆

230

2

130

1333

44 R

Q

R

QQVQW

πεπε

The potential energy stored in 3 charges is given by

( )332211

230

2

130

13

230

3

120

12

130

3

120

21

23

32

13

31

12

21

0

23

2

1

4444442

1

4

1

VQVQVQ

R

Q

R

QQ

R

Q

R

QQ

R

Q

R

QQ

R

QQ

R

QQ

R

QQWWW

++=

++

++

+=

++=∆+=

πεπεπεπεπεπε

πε

Note that V1, the potential at the position of Q1, is caused by charges Q2, Q3, and it is

different from the V1 in the two-charge case. Using the same procedure, the potential

energy (electrostatic energy) of a group of N discrete point charges can be given by

J)(2

1

1

∑=

=N

k

kke VQW

Likewise, the potential energy due to continuous charges can be given by

J)('2

1

'∫=V

ve VdvW ρ

Since the SI unit for energy, Joule (J), is too large, a more convenient unit, electron-

volt (eV), which is the energy or work required to move an electron against a potential

difference of one volt, i.e.,

Electrostatics Note

15

J)(1060.1 (eV)1 19−×= is used instead.

Example 3-17 Find the energy required to assemble a uniform sphere of charge of

radius b and volume charge density ρv.

3-10.1 Electrostatic Energy in Terms of Field Quantities

Since vρ=⋅∇ D , ∫ ⋅∇='

')(2

1

Ve VdvW D ,

using ( ) VVV ∇+⋅∇=⋅∇ DDD yields

∫∫∫∫ ⋅+⋅=∇−⋅∇=''''

'2

1'

2

1'

2

1')(

2

1

VSVVe dvdVVdvdvVW EDsDDD .

Since at least D is proportional to 1/r

2 and V is proportional to 1/r

, let V’ be the sphere

of radius r, and taking r→∞, the first term of the right hand side vanishes. Hence,

(J)'2

1

'∫ ⋅=V

e dvW ED

For a linear, isotropic medium,

(J)'2

1

'

2∫=V

e dvEW ε

Here, one can define electrostatic energy density we as

∫=='

32 '2

1;)(J/m

Veee dvwWEw ε

Example 3-18 A parallel-plate capacitor

Example 3-19 A cylindrical capacitor (Figure 3)

3-10.2 Electrostatic Forces

Recall that Coulomb’s law governs the force between two point charges, but it might

be hard to determine the force using Coulomb’s law in a more complex system of

charged bodies. In such cases, the following principle is useful.

Principle of Virtual Displacement : calculate the force on an object in a charged

system from the electrostatic energy of the system

Electrostatics Note

16

Consider an isolated system of charged conducting, as well as dielectric, bodies

separated from one another with no connection to the outside world. The mechanical

work done by the system to displace one of the bodies by a differential distance dl (a

virtual displacement) is given by

lF ddW Q ⋅= (3-112)

where FQ denotes the total electric force. Since it is an isolated system with no

external supply of energy, the mechanical work must be done at the expense of the

stored electrostatic energy, i.e.,

llllddWdW Qe ⋅=−= F . (3-113)

Writing the force in terms of the gradient of the work, one can write

lllldWdW ee ⋅∇= )( (3-114)

Since dl is arbitrary, comparison of (3-113) and (3-114) yields (N)eQ W−∇=F

Example 3-20 the force on conducting plates of a parallel-plate capacitor

3-11 SOLUTION OF ELECTROSTATIC BOUNDARY-VALUE PROBLEMS

So far, techniques for determining E, D, V, etc for a given charge distribution have

been discussed. In many practical problems, the charge distribution is not known

everywhere. In such cases, differential equations that govern the electric potential in

an electrostatic situation are formulated, and the boundary conditions are applied to

obtain what are called boundary-value problems.

3-11.1 Poisson’s and Laplace’s Equations

In Electrostatics,

Vv −∇==×∇=⋅∇ E0ED ;;ρ In linear, isotropic medium, since ED ε= ,

vV ρεε =∇−⋅∇=⋅∇=⋅∇ )()( ED

Hence, one obtains Poisson’s equation and Laplace’s equation (ρv =0 case) as

follows:

ερ vVsPoisson −=∇ 2:)'( ; 0:)'(

2 =∇ VsLaplace

where ∇2(del square) is called Laplacian operator.

3-11.1* Uniqueness Theorem

Uniqueness theorem asserts that a solution of an electrostatic problem satisfying its

boundary conditions (Poisson’s equation or Laplace’s equation) is the only possible

solution, irrespective of the method by which the solution is obtained.

Proof Suppose a volume τ is bounded outside by a surface So which may be a surface

at infinity. Inside the closed surface So there are a number of charged conducting

bodies with surfaces S1, S2, …, Sn at specified potentials, as depicted in Fig. 11.

Electrostatics Note

17

Figure 11 proof of

Uniqueness theorem

Now assume that, contrary to the uniqueness theorem,

there are two solutions, V1 and V2, to Poisson’s

equation in τ:

ερ

ερ vv VV −=∇−=∇ 2

2

1

2 ;

Also assume that both V1 and V2 satisfy the same

boundary conditions on S1, …, Sn and So. Let Vd= V1-

V2, then

nSSdd VV,,in in

2

1

0;0K

==∇τ

Using ( ) fff ∇+⋅∇=⋅∇ AAA and letting f=Vd, A=

∇Vd yields

( ) 222)()()( dddddddddd VVVVVVVVVV ∇=∇+∇=∇⋅∇+∇⋅∇=∇⋅∇

Integrating both sides of the equation above yields

( ) ( )∫ ∫∫ ∇=⋅∇=∇⋅∇S

ddddd dvVdVVdvVVττ

2s

Since Vd = 0 on S1, S2, …, Sn and on S0

22 ;/1;/1; rdsrVrVr dd ∝∝∇∝∞→ .

Thus, the integral on the left hand side vanishes. Since |∇Vd|2 is nonnegative, |∇Vd|

must be identically 0, which means Vd has the same value at all points in τ as it has on

the bounding surfaces, S1,…,Sn, where Vd=0. Thus, Vd=0 everywhere, and therefore

V1=V2, i.e., only one solution exists.

Example 3-21 parallel conducting plates separated by d with ρv=- ρ0y/d

Example 3-22 Two infinite insulated conducting plates maintained at potentials 0 and

V0 (Figure 12) Find the potential distribution for 0<φ <α and α <φ<2π.

Figure 12 Example 3-22

Example 3-23 Given the inner and outer radii of two concentric, thin, conducting,

spherical shells (Ri, Ro),

Electrostatics Note

18

Respectively, and the space between the shells is filled

with a dielectric. Determine the potential distribution in

the dielectric material by solving Laplace’s equation.

Example 3-16** A spherical capacitor

3-11.5 Method of Images

The method of images is the technique applied to boundary value problems by

replacing boundary surfaces with appropriate image charges, instead of attempting to

solve a Poisson’s or Laplace’s equation.

Example 3-24 Point Charges Near Conducting Planes as shown in Fig. 13 (a)

(a) Physical arrangement

(b) Image charge and field lines

Fig. 13

A formal procedure would require the solution of Poisson’s

equation in the y > 0 region with boundary conditions V = 0 at y =

0 and at infinity. Here, if an appropriate image charge can be used

to replace the conducting plane such that all boundary conditions

are satisfied, then the solution would be obtained in a

straightforward manner. Suppose one replaces the conductor with

the charge –Q at (0,-d,0), then the potential at a point P(x, y, z) is

given by

<

≥

+++−

+−+=

−=

−+

00

0)(

1

)(

1

4

11

4),,(

2222220

0

y

yzdyxzdyx

Q

RR

QzyxV

πε

πε

.

Note that the condition V=0 at y=0 is satisfied. Then, E for y≥0 is

given by

[ ] [ ]

+++

+++−

+−+

+−+=−∇=

2/32222/32220 )(

ˆ)(ˆˆ

)(

ˆ)(ˆˆ

4 zdyx

zdyx

zdyx

zdyxQV

zyxzyxE

πε

Hence, the surface charge density becomes ( ) 2/322200

2 zdx

QdE

yys

++−==

= περ .

Electrostatics Note

19

Line charge near a parallel conducting cylinder

Consider the problem of a line charge ρℓ located at a distance d from the axis of a

parallel, conducting, circular cylinder of radius a. Both are assumed to be infinitely

long. Fig. (a) shows a cross section of this arrangement. To apply the method of

images, first observe that (1) The image must be a parallel line charge inside the

cylinder in order to make the cylindrical surface at r = a an equipotential surface. Let

call this image line charge ρi (2) Because of the symmetry with respect to the line OP,

the image line charge must lie somewhere along OP, say at a point Pi, which is a

distance di from the axis (Fig. (b)). The unknowns needed to be determined here are ρi

and di. First, let l

ρρ −=i , then the potential at a distance r from a line charge of

density ρℓ is given by

r

rdr

rdrEV

r

r

r

rr

0

00

ln2

1

2 00 περ

περ

ll ∫∫ =−=−=

Thus, the potential at point M can be found by adding contributions of ρℓ and ρi, i.e.,

r

r

r

r

r

rV i

i

M ln2

ln2

ln2 0

0

0

0

0 περ

περ

περ

lll =−=

In order for an equipotential surface to coincide with the surface r=a, ri/r must be a

constant. The point Pi must be located such that ∆OMPi is similar to ∆OPM, i.e., ∠

OMPi=∠ OPM. Hence,

constantor; =====d

a

a

d

r

r

OP

OM

OM

OP

PM

MP iiii

Therefore, di=a

2/d . The point Pi is called inverse point of P with respect to a circle of

radius a.

Example 3-25 Capacitance per unit length between two long parallel circular

conducting wires of radius a